POJ 3630 Phone List(trie 首碼)

Phone List Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 19045   Accepted: 6016 Description Given a list of phone numbers, determine if it is consistent in the sense that no number is the prefix of another. Let's say the phone catalogue listed these numbers: Emergency 911 Alice 97 625 999 Bob 91 12 54 26 In this case, it's not possible to call Bob, because the central would direct your call to the emergency line as soon as you had dialled the first three digits of Bob's phone number. So this list would not be consistent. Input The first line of input gives a single integer, 1 ≤ t ≤ 40, the number of test cases. Each test case starts with n, the number of phone numbers, on a separate line, 1 ≤ n ≤ 10000. Then follows n lines with one unique phone number on each line. A phone number is a sequence of at most ten digits. Output For each test case, output "YES" if the list is consistent, or "NO" otherwise. Sample Input 2391197625999911254265113123401234401234598346 Sample Output NOYES Source Nordic 2007

這個題目典型的trie樹求首碼，注意後面插入的只要是前面走過的路徑就是重複

就是沒有任何節點更新！

#include <iostream>#include <stdio.h>#include <string.h>using namespace std;struct trie{trie *next[10];int num;}re_root,po[100000];int pos;int init(trie *root){memset(root->next,0,sizeof(root->next));root->num=0;return 0;}int insert(trie *root,char *name){if(root->num > 0)return 0;if(name[0]==0){if(root->num > 0)return 0;elseroot->num=1;return 1;}if(root->next[name[0]-'0']){if(name[1]==0)return 0;return insert(root->next[name[0]-'0'],name+1);}else{root->next[name[0]-'0']=&po[pos];init(&po[pos]);pos++;return insert(root->next[name[0]-'0'],name+1);}return 1;}int main(){int n;int i;char str[230];bool flag;int t;scanf("%d",&t);while(t--){scanf("%d",&n);flag=true;pos=0;init(&re_root);for(i=0;i<n;i++){scanf("%s",str);if(flag&&insert(&re_root,str)==0)flag=false;}if(flag)printf("YES\n");elseprintf("NO\n");}return 0;}